The maximum value of f(x) = displaystylex/(x^2 + 9) , x ∈ R , is

The correct answer is Option (1) → \(\displaystyle\frac{1}{6}\)

By the quotient rule:

    \( \displaystyle f'(x) = \frac{(x^{2} + 9) - x \cdot 2x}{(x^{2} + 9)^{2}} = \frac{9 - x^{2}}{(x^{2} + 9)^{2}} \)

The sign of f′(x) is the sign of \(9 - x^{2}\):

    \(f'(x) \gt 0\) for \(-3 \lt x \lt 3\)
    \(f'(x) \lt 0\) for \(x \gt 3\) and for \(x \lt -3\)

So f rises up to x = 3 and falls after it (and x = −3 gives a minimum). By the first derivative test the maximum value is

    \( \displaystyle f(3) = \frac{3}{9 + 9} = \frac{1}{6} \)

For x ≤ 0, f(x) ≤ 0, so no value beats \(\frac{1}{6}\).